INDETERMINATE CUBIC EQUATIONS 57 



For if in (6) we replace >., ^, a, b by XyV, \LJV, a/c, b/c respec- 

 tively we shall obtain for (7) that if 



\a?+v&=vc*, (8) 



then xZ 3 + [ *7 3 =vZ 3 , 



where Z=a([x6 3 +vc 3 ), 7=-6(Xa 3 +vc 3 ), Z=c(Xa 3 -{x6 3 ). (9) 



For example, if the equation were 



then a= 1, 6=1, c=l ; X=5, ^=6, v=l, and consequently 



X=-7, 7=4, Z=-ll. 



From this solution we may proceed to derive a third, and so on 

 ad infinitum. 



Cor. If in (8) we put X =(/.=!/, we derive that if 



then 1 a 3 (6 3 +c 3 ) 3 -6 3 (c 3 +a 3 ) 3 =c 3 ( 3 -6 3 ) 3 . 



QUESTION 6. Express 5, 17, and 41 algebraically, each as 

 the sum of five integral cubes. 2 



Taking the case of 5 first, since 5=5 3 +2 3 4 3 4 3 O 3 , if we 

 assume 



5=(f-4) 3 +(af-4) 3 +(-af+2) 3 +(6a 3 +(-^+5) 3 (1) 

 it is clear that, on expansion, the coefficient of f 3 will be 

 unity, while the constant term disappears. If, therefore, we 

 can make the coefficient of vanish by expressing both a and 

 b integrally in terms of some unknown, then we shall obtain 

 an integral algebraical value for f which will render (1) an 

 identity. Now (1) on expansion is 



?+3(-4-2a a +56 2 )f 2 +3(16+12a-25&)f=0. (2) 



Hence to make the coefficient of vanish we must have 

 256=12a+16, and this is done integrally by taking a=25H-7, 

 6=12^+4. Substituting these values in (2), we find for the 

 required value of 



1590* 2 +660Z+66. 



1 This result is due to Tait. See Chrystal's Algebra, part i. chap, xiv., Ex. xx. 

 No. 2. 



2 Cf . Mathematics from the Educational Times, New Series, vol. iv. No. 15225. 



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